Question

What is the least multiple of 23 which when divided by 18,21 and 24 leaves remainders 7, 10
and 13 respectively?​

Answer 1

Answer:

With 18 as the devisor, the required remainder is 7, with 21 as the divisor, the required remainder is 10 and with 24 as the remainder, the required remainder is 13. In modulo arithmatic, all these (7, 10 and 13) are equivalent to -11 because 7–18 = 10–21=13–24 = -11.

So, the first part is easy, we have to find the LCM of the divisors i. e. 18, 21 and 24. The LCM of 18, 21 and 24 is 504 and if we add -11 to it (i. e. subtract 11 from it), we get 493 and this is the required number that leaves the remainders 7, 10 and 13 with the divisors 18, 21 and 24 respectively.

But, we also want such a number to be completely divisoble by 23. So here comes the second part of the solution.

The number that is 11 short of some multiple of 504 (LCM) is completely divisible by 23. So, the equation becomes:

504*X - 11 = 23*Y

Now, we have 2 variables and there’s only 1 equation, so obviously we are going to have infinite solutions and we will have to proceed with trial and error method to obtain the first solution.

If we try with X as 1, we get a number that leaves 10 as the remainder with 23. If we try with X as 2, we get a number that leaves 8 as the remainder with 23. So we expect 0 as the remainder when X = 6.

If we try with X as 6, indeed we get 0 as the remainder with the number 504*6–11 = 3013.

Thus, you see, 3013 is the smallest number that is completely divisible by 23 and at the same time leaves remainders 7, 10 and 13 with divisors 18, 21 and 24 respectively.

Step-by-step explanation: